UPSC NDA I 2023 Mathematics Solved Paper

Concept:Using the sine rule and given side-angle relation to find the angles of triangle ABC.Explanation:Given: $a = \sqrt{3}$, $b = 1$, and the relation $a + b + c = 6\left(\frac{\sin A + \sin B + \sin C}{3}\right)$ leads to $a + b + c = 2(\sin A + \sin B + \sin C)$.By the sine rule: $\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = k$.So $\sin A = ak$, $\sin B = bk$, $\sin C = ck$.Substitute into the relation: $a + b + c = 2(ak + bk + ck) = 2k(a + b + c)$.Assuming $a+b+c \neq 0$, we get $k = \frac{1}{2}$.Now $\sin A = a k = \sqrt{3} \times \frac{1}{2} = \frac{\sqrt{3}}{2}$ → $A = 60^\circ$.$\sin B = b k = 1 \times \frac{1}{2} = \frac{1}{2}$ → $B = 30^\circ$.Using $A + B + C = 180^\circ$: $60^\circ + 30^\circ + C = 180^\circ$ → $C = 90^\circ$.Thus, $\angle C = 90^\circ$ so triangle is right-angled (Statement 1 true).The angles $30^\circ, 60^\circ, 90^\circ$ are in arithmetic progression (common difference $30^\circ$), so Statement 2 is true.Answer:Both statements 1 and 2 are correct. Option C.